$-10de + df + 2d - 1 = -e + 3$ Solve for $d$.
Combine constant terms on the right. $-10de + df + 2d - {1} = -e + {3}$ $-10de + df + 2d = -e + {4}$ Notice that all the terms on the left-hand side of the equation have $d$ in them. $-10{d}e + 1{d}f + 2{d} = -e + 4$ Factor out the $d$ ${d} \cdot \left( -10e + f + 2 \right) = -e + 4$ Isolate the $d$ $d \cdot \left( -{10e + f + 2} \right) = -e + 4$ $d = \dfrac{ -e + 4 }{ -{10e + f + 2} }$